Two blocks of masses `m_(1)` and `m_(2)` are connected with a light spring of force constant k and the whole system is kept on a frictionless horizontal surface. The masses are applied forces `F_(1)` and `F_(2)` as shown in fig. At any time the blocks have same acceleration `a_(0)` but in opposite direction. Now answer the following :

The value of `a_(0)` is

Options

A. `(F_1)/(m_1)-(m_(2)g)/(m_(1))=(F_2)/(m_2)-(m_(1)g)/(m_2)`
B. `(F_1)/(m_2)=(F_2)/(m_1)`
C. `(F_1)/(m_1)+(m_(2)g)/(m_(1))=(F_2)/(m_2)+(m_(1)g)/(m_2)`
D. `(F_1)/(m_1)=(F_2)/(m_2)`

Answer

D

Explanation

In horizontal direction, both `m_(1)` and `m_(2)` will have same acceleration for any case. In vertical direction also, they should have same acceleration. Let it be a upwards for boths, then
`F_(1)-m_(1)g=m_(1)a`..(i)
`F_(2)-m_(2)g=m_(2)a` …(ii)
From (i) and (ii) , `(F_(1))/(m_(1))=(F_(2))/(m_(2))`.